Motion of m1:
The acceleration is a0 in the horizontal direction. The forces on m1 are:
1) T by the string (horizontal)
2) m1g by the earth (vertically downwards) and
3) N by the table (vertically upwards)
In the horizontal direction the equation is,
T = m1a0 -------------------- (2)
Motion of m2:
Acceleration is a0-a in the downward direction, the forces on m2 are
1) m2g downward by the earth and
2) T = T/2 upward by the string
Thus 
----------------------- (3)
Motion of m3:
Acceleration is (a0+a) downward, the forces on m3 are
1) m3g downward by the earth and
2) T1 = T/2 upward by the string thus,
To calculate a0 eliminate T and a from (2), (3) and (4)
5) As shown in figure (38) the wedge is fixed to the ground the prism B of mass ‘M’ and the block C of mass ‘m’ is placed as shown. Find the acceleration of the block C w.r.t B when the system is set free, Neglect any friction.
Fig (38)
Solution:
Let ‘a’ be the acceleration of B towards, then
F.B.D of C relative to B is as shown in fig (a).
Þ Mg – N-maSina = 0 as C is always in contact with B.
Þ N = mg-maSina --------------------- (1)
And maCosa = ma1 where a1 = acceleration of C relative to B
Þ a1 = aCosa -------------------- (2)
F.B.D of B (As shown in fig (b))
Þ MgSina + NSina = Ma
Putting values of N from (1) we get
Þ MgSina +mgSina - maSin2a = Ma ----------------------- (3)
Dumb Question:
1) Why doesn’t the block move w.r.t B in fig (a)?
Ans: Suppose instead of C, you are standing on B. Now you will see that w.r.t B, you don’t move in vertical direction as neither can you ‘fly’ up from ‘B’ nor can you press down so much that you move down.
6) Two blocks A and B of masses m1 and m2 respectively are placed on each other and their combination rest on a fixed horizontal surface. A mass less string passing over a smooth pulley as shown in figure (39) is used to connect A and B. Assuming the coefficient of sliding friction between all surfaces to be m. Show that both A and B will move with a uniform speed. If A is dragged with a force F = m (3m1+m2) g to left.
Fig (39)
Solution:
For vertical equilibrium,
N1-m1g = 0
Þ N1 = mg ----------------------- (1)
As m1 slides on m2 hence
f1= mN1 = mm1g --------------------- (2)
For horizontal equilibrium,
F-T-f1 = 0
Þ F-T-mm1g = 0 ---------------------- (3)
For vertical equilibrium of m1
N2 – N1-m2g = 0
Þ N2 = N1+ m2g = m1g + m2g ------------------------- (4)
Since f2 = mN2 = m (m1 + m2) g ------------------------ (5)
For horizontal equilibrium,
f1 + f2 – T = 0
Þ mm1g + m (m1+m2) g – T = 0 ---------------------- (6)
Subtracting (3) from (6) we get
3mm1g + mm2g – F = 0
Þ F = m (3m1 + m2) g.
7) Figure (40) shows a small block A of mass m kept at the left end of a plank B of mass M =2m and length l, the system can slide on a horizontal road the system is started towards right with initial velocity v the friction coefficient between the road and the plank is ½ and that between the plank and the block is ¼ find:
1) The time elapsed before the block separates from the plank.
2) Displacement of block and plank relative to ground till that moment.
Fig (40)
Solution:
There will be relative motion between block and plank and plank and road. So at each surface limiting friction will act the direction of friction forces at different surfaces are as shown in figure.
Here 
and 
Retardation of A is
And retardation of B is
Relative acceleration of A with respect to B is
Initial velocity of both A and B is v, so there is no relative initial velocity hence.
1) Applying 
2) Displacement of block:
Displacement of plank:
- Tip: The problem, as you see, it becomes simpler using relative concept otherwise the problem is a bit complex!
